Bazza's 3/4 Question Thread

[WhoTookMyUsername]

1) in substitution, is the step du=dx or something similar, e.g. du=2dx
Mathemtatically notationally acceptable?
2) if the previous part asks for 4 dp, but we have a more detailed answer to 5 or 6dp, should we use this for the next question?

Thanks!

[BubbleWrapMan]

1) Not in my opinion, but you can probably get away with it in VCAA exams.

2) They usually don't put you in that situation, but you should use the value from the previous question.

[StumbleBum]

1) in substitution, is the step du=dx or something similar, e.g. du=2dx
Mathemtatically notationally acceptable?

My teacher said that it is not correct mathematically, can't remember why... In an exam i would just stick with du/dx=2 and then transpose it in your head before using it when you substitute. Although this does allow a risk for a very silly error! Maybe just write it then rub it out, although i doubt they would take any working marks off for that if you got the question wrong.

[WhoTookMyUsername]

Just confirming, but arg(z) is inversetan(re(z)/im(z)) and not (necessarily) inverstan(y/x))?

Also what are some methods of sketching the graph defined by
Arg(iz)<pi/4 ?
I converted it to cartesian, but was unsure but which side to shade, and which half of the cartesian to graph, had to sort of use intuition and transformations to decide, any better way? (Or alternative ways about thinking of the question?), how would you sketch this graph?
Thanks

[brightsky]

i'd suggest you always draw the triangle and work out the angle from scratch without appealing to any formulas. (and note that it's definitely not arctan(re(z)/im(z)) because tan = opp/adj not adj/opp).

consider i*z. it's z but rotated 90 degrees anticlockwise.

[WhoTookMyUsername]

Sorry yeah i meant im(z)/re(z)
Sometimes cannot draw a triangle though, like for this one, i can't draw the triangle because i can't draw the graph without that rule :/ (without intuition)


How do you intuitively know that though?
Like what process did you go through to graph arg(iz)<pi/4?
Or what is your thought process / is there any way to do it mathematically on the page and work it out rather than from intuition?
Like i'd prefer a mathmatical process that always works for this type of thing, less risky in the exam


I didsome dodgy thing like
Arg(iz) = pi/4
Then iz=cis(pi/4)
Z = cis(pi/4 - pi/2)
Arg(z) = -pi/4

I know that's mathematically all over the place though i *think* my reasoning is sound
Just looking for something similar to that where i can apply a certain technique to questions like this (or a way to tidy what i did up)

[brightsky]

"like for this one" - dont' see anything...but arctan(y/x) is exactly the same as arctan(imz/rez). this doesn't always give you the angle you want. you might need to do some adjusting depending on what quadrant the number is in.

and for Arg(iz), why not write Arg(iz) = Arg(z) + pi/2 and go from there?

[WhoTookMyUsername]

No it's not, iz becomes xi - y

I'm 99% sure Arg(iz) does not = Arg(z) + pi/2

IIRC
Arg(i+z) = Arg(i)Arg(z)
arg(a+b)=Arg(a)Arg(b)
cis(a+b)=cis(a)cis(b)


Still not sure how to go about graphing Arg(iz) < pi/4 without 'intuition' xD


Arg(iz) = pi/4
Then iz=cis(pi/4)
Z = cis(pi/4 - pi/2)
Arg(z) = -pi/4
Is closeish? Anyway to tidy :X

[rife168]

No it's not, iz becomes xi - y

I'm 99% sure Arg(iz) does not = Arg(z) + pi/2

IIRC
Arg(i+z) = Arg(i)Arg(z)
arg(a+b)=Arg(a)Arg(b)
cis(a+b)=cis(a)cis(b)


Still not sure how to go about graphing Arg(iz) < pi/4 without 'intuition' xD


Arg(iz) = pi/4
Then iz=cis(pi/4)
Z = cis(pi/4 - pi/2)
Arg(z) = -pi/4
Is closeish? Anyway to tidy :X

I think you have it the wrong way around...
Arg(ab)=Arg(a)+Arg(b)

[WhoTookMyUsername]

Yes i do.

Let z1=cis(a)
Let z2=cis(b)
Z1z2=cis(a)cis(b)
=cis(a+b)

Arg(z1z2) = a+b = arg(z1) + arg (z2)

Fml... Gotta go back and look over this stuff

[rife168]

Yes i do.

Let z1=cis(a)
Let z2=cis(b)
Z1z2=cis(a)cis(b)
=cis(a+b)

Arg(z1z2) = a+b = arg(z1) + arg (z2)

Fml... Gotta go back and look over this stuff

Well, it's better that you found that out now rather than after the exam...

[d3stiny]

Why not..

Arg(iz) = pi/4
arctan(Im(iz)/Re(iz)) = pi/4
Im(iz)/Re(iz) = 1 
and iz = xi - y
x / -y = 1
y = -x   for x > 0

Not sure if thats right though.

[WhoTookMyUsername]

Why not..

Arg(iz) = pi/4
arctan(Im(iz)/Re(iz)) = pi/4
Im(iz)/Re(iz) = 1 
and iz = xi - y
x / -y = 1
y = -x   for x > 0

Not sure if thats right though.

Yeah that's sorta what i did the first time, but there's two problems with that,
Firstly how do you know for sure that x>0 (not a rhetorical question, i'm asking lol)
And secondly, how do you know which region to shade in?

(Arg(iz)<pi/4)

Yes i do.

Let z1=cis(a)
Let z2=cis(b)
Z1z2=cis(a)cis(b)
=cis(a+b)

Arg(z1z2) = a+b = arg(z1) + arg (z2)

Fml... Gotta go back and look over this stuff

Well, it's better that you found that out now rather than after the exam...

Haha yeah it is, i must have overlooked something... This has never troubled me in any practice exams though D:
Thanks! And thanks brightsky

[WhoTookMyUsername]

What is the domain of x=cos(y)?
Is it R or [-1,1]?

I thought it would be R, but DHA had the domain as -1,1 (because he sketched it on the same graph as it's inverse)

But doesn't domain refer to the variable it's in terms of?
Like if it was m=cos(p)
The domain would be R, isn't it the same thing?

2)
 Is piSx^2 dy an acceptable line of working? (Notationally wise)
 Thanks

[Somye]

for 1, no, because in your other example (m = cos(p)), you label your vertical axis as m and horizontal as p

whereas in the question given, you cant really have x as your vertical axis, and so you have to use inverse etc.